Solve the following differential equations :
Find the particular solution of : `(dy)/(dx)=cos(x+y+2)`, given that `x=0, y=-2`.

Answer

To solve the differential equation \(\frac{dy}{dx} = \cos(x + y + 2)\) with the initial condition \(x = 0\) and \(y = -2\), we will follow these steps:

### Step 1: Substitute \(v = x + y + 2\)

Let \(v = x + y + 2\). Then we can express \(y\) in terms of \(v\) and \(x\):
\[ y = v - x - 2 \]

### Step 2: Differentiate \(v\) with respect to \(x\)

Now, differentiate \(v\) with respect to \(x\):
\[ \frac{dv}{dx} = \frac{d}{dx}(x + y + 2) = 1 + \frac{dy}{dx} \]

### Step 3: Substitute \(\frac{dy}{dx}\) in terms of \(\frac{dv}{dx}\)

From the previous step, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{dv}{dx} - 1 \]

### Step 4: Substitute into the original equation

Now substitute this expression into the original differential equation:
\[ \frac{dv}{dx} - 1 = \cos(v) \]
Rearranging gives:
\[ \frac{dv}{dx} = \cos(v) + 1 \]

### Step 5: Separate variables

Now we can separate the variables:
\[ \frac{dv}{\cos(v) + 1} = dx \]

### Step 6: Integrate both sides

Integrate both sides:
\[ \int \frac{dv}{\cos(v) + 1} = \int dx \]

The integral on the left can be simplified using the identity \(\cos(v) + 1 = 2\cos^2(v/2)\):
\[ \int \frac{dv}{\cos(v) + 1} = \int \frac{dv}{2\cos^2(v/2)} = \int \frac{1}{2} \sec^2(v/2) dv = \tan(v/2) + C \]
Thus, we have:
\[ \tan\left(\frac{v}{2}\right) = x + C \]

### Step 7: Substitute back for \(v\)

Recall that \(v = x + y + 2\):
\[ \tan\left(\frac{x + y + 2}{2}\right) = x + C \]

### Step 8: Use the initial condition to find \(C\)

Now we apply the initial condition \(x = 0\) and \(y = -2\):
\[ \tan\left(\frac{0 + (-2) + 2}{2}\right) = 0 + C \implies \tan(0) = C \implies C = 0 \]

### Step 9: Write the particular solution

Thus, the particular solution is:
\[ \tan\left(\frac{x + y + 2}{2}\right) = x \]

### Final Step: Solve for \(y\)

To express \(y\) explicitly, we can rearrange:
\[ \frac{x + y + 2}{2} = \tan^{-1}(x) \implies x + y + 2 = 2\tan^{-1}(x) \implies y = 2\tan^{-1}(x) - x - 2 \]

### Final Answer

The particular solution to the differential equation is:
\[ y = 2\tan^{-1}(x) - x - 2 \]